Question

# Let f and g be real valued functions such that f(x+y)+f(x-y)=2f(x).g(y)$$\forall x, y\epsilon R$$. Prove that, if f(x) is not identically zero and $$\left | f\left ( x \right ) \right |\leq 1\forall x\epsilon R$$, then $$\left | g\left ( y \right ) \right |\leq 1\forall y\epsilon R$$.

Solution

## Let maximum value of $$f(x)$$ be $$M$$.$$\Rightarrow max|f(x)|=M,$$ where $$0< M\leq 1$$          (i)(Since, $$f$$ is not identically zero and $$\left | f\left ( x \right ) \right |\leq 1\forall x\epsilon R$$)Now, $$f(x+y)+f(x-y)=2f(x).g(y)$$$$\Rightarrow$$  $$|2f(x)|.|g(y)|=|f(x+y)+f(x-y)|$$$$\Rightarrow$$   $$2\left | f\left ( x \right ) \right |.\left | g\left ( y \right ) \right |\leq \left | f\left ( x+y \right ) \right |+\left | f\left ( x-y \right ) \right |$$          (as $$\left | a+b \right |\leq \left | a \right |+\left | b \right |$$)$$\Rightarrow$$   $$2\left | f\left ( x \right ) \right |.\left | g\left ( y \right ) \right |\leq M+M$$          [Using Eq. (i), ie, $$max|f(x)|=M$$]$$\Rightarrow$$   $$\left | g\left ( y \right ) \right |\leq 1$$ for $$y\epsilon R$$Thus, if $$f(x)$$ is not identically zero and $$\left | f\left ( x \right ) \right |\leq 1\forall x\epsilon R$$then, $$\left | g\left ( y \right ) \right |\leq 1$$ for $$y\epsilon R$$Mathematics

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